Question

Two additional jurors are needed to complete a jury for a criminal trial. There are six prospective jurors, two women and four men. Two jurors are randomly selected from the six available. Let Y represent the number of women in this group of two additional Jurors.

a) Find the probability distribution of Y.
b) E(Y).
c) V(Y).

Answer 1

The probability distribution of Y, the number of women selected from the group of two additional jurors, is: P(Y=0) = 2/5, P(Y=1) = 12/15, P(Y=2) = 1/15. The expected value of Y is 14/15. The variance of Y is 16/25.

Step-by-step explanation:

To find the probability distribution of Y, we can consider all possible combinations of selecting two jurors from the six available. Since there are two women and four men, the possible values for Y are 0, 1, and 2. The probability distribution of Y is as follows:

P(Y=0) = P(selecting two men) = (4/6) * (3/5) = 2/5

P(Y=1) = P(selecting one man and one woman) = (4/6) * (2/5) + (2/6) * (4/5) = 4/15 + 8/15 = 12/15

P(Y=2) = P(selecting two women) = (2/6) * (1/5) = 1/15

b) To find E(Y), we can multiply each value of Y by its corresponding probability and sum them up:

E(Y) = 0 * (2/5) + 1 * (12/15) + 2 * (1/15) = 0 + 12/15 + 2/15 = 14/15

c) To find V(Y), we can use the formula V(Y) = E(Y^2) - (E(Y))^2. Since we already know E(Y), we need to calculate E(Y^2):

E(Y^2) = (0^2 * (2/5)) + (1^2 * (12/15)) + (2^2 * (1/15)) = 0 + 12/15 + 4/15 = 16/15

V(Y) = E(Y^2) - (E(Y))^2 = (16/15) - (14/15)^2 = 16/15 - 196/225 = 144/225 = 16/25